If 3(cot²θ - cos²θ) = cos²θ, 0°< θ < 90°, then the value of (tan²θ + cosec²θ + sin²θ) is:

Solution (By Examveda Team)

$$\eqalign{ & 3\left( {{{\cot }^2}\theta - {{\cos }^2}\theta } \right) = {\cos ^2}\theta \cr & 3{\cot ^2}\theta = {\cos ^2}\theta + 3{\cos ^2}\theta \cr & 3{\cot ^2}\theta = 4{\cos ^2}\theta \cr & \frac{3}{4} = {\sin ^2}\theta \cr & \sin \theta = \sin {60^ \circ } \cr & \theta = {60^ \circ } \cr & {\tan ^2}\theta + {\text{cose}}{{\text{c}}^2}\theta + {\sin ^2}\theta \cr & = {\tan ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & = 3 + \frac{4}{3} + \frac{3}{4} \cr & = \frac{{36 + 16 + 9}}{{12}} \cr & = \frac{{61}}{{12}} \cr} $$